Homework Help: Another vector problem -- A ball suspended from 3 wires

1. Nov 17, 2014

max1995

Question

A ball with a mass 4 kg is suspended from 3 light in-extensible wires (diagram given)
Given that the ball is in equilibrium, calculate the tension in each of the wires.

My attempt

T1 is equal to the weight of the ball which is 4g
so T1 = 4g

From there I dont know what to do but for T3 is the vertical component equal to 4g and to find T3 you can do 4gcos(70)= T3?

is the above for T3 right? If not how do I do it?

Not sure how to do T2 could someone please explain?

Thanks

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2. Nov 17, 2014

Vagn

Work with the horizontal components first. We know the horizontal components of T2 and T3 must be equal in magnitude. You can then relate T2 and T3.

3. Nov 17, 2014

Simon Bridge

You know the angles the tension forces make to each other, and you know one magnitude.
You know how to add vectors head-to-tail ... and you know what shape these vectors have to make when you do that.
You know some rules for finding lengths of sides when you know the angles.
(Or split everything into horizontal and vertical components...)

4. Nov 17, 2014

max1995

Thank you for the help :)

so horizontal of T2=T2cos60
horizontal of T3= T3sin70
so T2cos60=T3sin70 (is this right?)

5. Nov 17, 2014

Staff: Mentor

OK so far. Now, can you come up with a similar equation using the vertical components of T1, T2 and T3?

6. Nov 17, 2014

max1995

T3cos(70)= vertical of T3
T2sin(60)= vertical of T2
T1=4g

Does this mean that for the verticals of the tensions 4g=T3cos(70) + T2sin(60)?

7. Nov 17, 2014

Staff: Mentor

Not quite. T2 is upwards. T1 and T3 are downwards.

8. Nov 17, 2014

max1995

Okay so the verticals of T1 and T3 must equal the vertical of T2?

9. Nov 17, 2014

Staff: Mentor

Correct.

With horizontal components, the total to the left equals the total to the right.
With vertical components, the total upwards equals the total downwards.

(for a system in equlibrium, like this one.)

10. Nov 17, 2014

max1995

I dont understand what to do after that because substituting in the equations means the tensions cancel out of the equation?

11. Nov 17, 2014

Staff: Mentor

You have two equations in two unknowns, T2 and T3. They should not cancel out when you solve the equations together. If they do, you've made a mistake in your algebra. Can you show your work?

12. Nov 17, 2014

max1995

Starting equations:
T2cos(60)=T3sin(70)
T2(sin60) = T3cos(70) + 4g

Then do I rearrange one of them to make one of the T's the subject and sub it into the other equation?

Thank you for all the help btw :)

13. Nov 17, 2014

Staff: Mentor

Exactly. :D

14. Nov 17, 2014

max1995

Would you be able to give me an example of how to rearrange and substitute these equations to find one of the tensions? as im having trouble doing this

15. Nov 17, 2014

Staff: Mentor

Show us what you've tried and where you get "stuck" and I or someone else can probably help you get "unstuck."

16. Nov 17, 2014

max1995

Starting equations:
T2cos(60)=T3sin(70) (Equation 1)
T2(sin60) = T3cos(70) + 4g (Equation 2)

Rearranging T2(sin60) = T3cos(70) + 4g to get T2 = (T3cos(70) + 4g)/sin(60) then substitue it into equation 1

this gives (T3cos(70) + 4g)/sin(60)xcos(60) = T3sin(70)

then divided both sides by cos(60) to give

(T3cos(70) + 4g)/sin(60) = (T3sin(70))/cos60

Times both sides by (sin60)

(T3cos(70) + 4g = [(T3sin(70))/cos60]x sin60

this gives in numerical form

0.3420201433T3 + 39.24 = 1.627595363T3

Then subtract 0.3420201433T3 from both sides

39.24 = 1.28557522T3

then divde both sides by 1.28557522

T3 = 30.52N

for T3 this is what I got, is it correct?

17. Nov 17, 2014

Staff: Mentor

I'm starting to look over your algebra, but you can check the final answers yourself. Use your value of T3 to find T2, then substitute both into your starting equations (at the top of your post) and see if they work out.

18. Nov 17, 2014

Staff: Mentor

It looks OK to me so far. I wasn't sure at first because in your first post I missed the mass of the ball being "4 kg" and assumed that by "4g" in your equations you meant 4 grams. So I got a different number for T3 at first. I fixed that and now we agree.

19. Nov 17, 2014

max1995

yeah sorry i mean 4 times by gravity when saying 4g
Okay thank you for all the help :D

to find T2 is it alright just to sub in my value of T3 into one of my first equations to get its value?

Also is there a rep system on here so I can rep you? (new to this forum)

20. Nov 17, 2014

Staff: Mentor

Right. Either equation should work. If you try both of them and get the same value for T2, that will also serve as a check.

Not the kind that I think you're thinking of. We have other ways of identifying the "good guys."

21. Nov 17, 2014

Simon Bridge

There's a "like" button below threads.

I like the geometric approach.

You have $\vec T_1 + \vec T_2 +\vec T_3 = 0$
Draw those head-to-tail and you get a triangle.
Sketch out a scalene triangle but the $T_1$ side is vertical - try to get the directions approximately right.
Label the triangle like you would in maths class: the sides are a-b-c going clockwise around from $T_1$
The angles opposite the sides are labelled with the capitol letters so A is opposite a etc.

So:
$a=T_1=mg,\; b=T_2,\; c=T_3$ ... easy.
The angles have to be worked out carefully ...
c makes 70deg to the vertical, so the external angle between c and a is 70deg, so B=180-70=110
b makes 60deg to the horizontal, which is 150deg to the vertical, so the exterior angle between a and b is 150deg, so C=180-150=30deg
That means that A=180-110-30=40deg
Now you can use the sine rule to find $T_2$ and $T_3$

You'd actually end up with the same equatons - only faster with less chance of making an algebraic slipup.